A fair die is rolled 6,000 times. Assume the trials are independent and the probability of sucess = 1/6 of each trial. What's the approximate probability of getting at least 1050 fours in your 6,000 rolls of the die?
A. 044
Β. .048
C. 035
D..05
Ε. 063
To solve this problem, we can use the normal approximation to the binomial distribution since the number of trials is large and the probability of success is small.
The mean of the number of fours rolled in 6,000 trials is given by:
Mean = n * p = 6000 * (1/6) = 1000
The standard deviation of the number of fours rolled in 6,000 trials is given by:
Standard Deviation = sqrt(n * p * (1 - p)) = sqrt(6000 * (1/6) * (5/6)) ≈ 19.365
Now, we need to find the z-score for 1050 fours:
z = (1050 - 1000) / 19.365 ≈ 2.581
Using a standard normal distribution table or calculator, we find that the probability of getting at least 1050 fours is approximately 0.995, which is the area to the right of the z-score of 2.581.
Therefore, the approximate probability of getting at least 1050 fours in 6,000 rolls of the die is 1 - 0.995 = 0.005, which is equivalent to 0.5%.
So, none of the given options accurately represent the probability calculated above.